# IIFT 2012 QA | Previous Year IIFT Paper

**1. IIFT 2012 QA | Algebra - Number Theory**

If k is an integer and 0.0010101 × 10^{k} is greater than 1000, what is the least possible value of k?

- A.
4

- B.
5

- C.
6

- D.
7

Answer: Option C

**Explanation** :

We need 0.0010101 × 10^{k}^{ }> 1000

∴ 1010.1 × 10^{–6} × 10* ^{k}* > 1000

∴ 1010.1 × 10^{–6 + k} > 1000

∵ 1010.1 > 1000, 10^{–6 + k} = 1

∴ *k* – 6 = 0

∴ *k* = 6

Hence, option (c).

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**2. IIFT 2012 QA | Modern Math - Probability**

Ashish is studying late into the night and is hungry. He opens his mother’s snack cupboard without switching on the lights, knowing that his mother has kept 10 packets of chips and biscuits in the cupboard. He pulls out 3 packets from the cupboard, and all of them turn out to be chips. What is the probability that the snack cupboard contains 1 packet of biscuits and 9 packets of chips?

- A.
$\frac{6}{55}$

- B.
$\frac{12}{73}$

- C.
$\frac{14}{55}$

- D.
$\frac{7}{50}$

Answer: Option C

**Explanation** :

There are at least three packets of chips in the cupboard. There are 10 packets in all.

∴ (Number of packets of chips, Number of packets of biscuits) ≡ (3, 7) or (4, 6) or (5, 5) or (6, 4) or (7, 3) or (8, 2) or (9, 1) or (10, 0)

The number of ways in which three packets of chips can be drawn ^{3}C_{3} + ^{4}C_{3} + ^{5}C_{3} + ^{6}C_{3} + ^{7}C_{3} + ^{8}C_{3} + ^{9}C_{3} + ^{10}C_{3} = 330

The number of ways in which three packets of chips can be drawn when there are 9 packets of chips = ^{9}C_{3} = 84

∴ Required probability = $\frac{84}{330}$ = $\frac{14}{55}$

Hence, option (c).

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**3. IIFT 2012 QA | Algebra - Simple Equations**

The equation 7x – 1 + 11x – 1 = 170 has

- A.
no solution

- B.
one solution

- C.
two solutions

- D.
three solutions

Answer: Option B

**Explanation** :

7^{x}^{ – 1} + 11^{x}^{ – 1} = 170

WE can see that the RHS is a multiple of 10.

11^{x}^{ – 1} has 1 in its units place.

∴ 7^{x}^{ – 1} should have 9 in its units place.

The lowest value for which this is true is *x* = 3

7^{2} + 11^{2} = 170

We can see that the for any other value of *x*, which is greater than 3,

7^{x }^{– 1} + 11^{x}^{ – 1} > 170

Hence, option (b).

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**4. IIFT 2012 QA | Arithmetic - Percentage**

The annual production in cement industry is subject to business cycles. The production increases for two consecutive years consistently by 18% and decreases by 12% in the third year. Again in the next two years, it increases by 18% each year and decreases by 12% in the third year. Talking 2008 as the base year, what will be the approximate effect on cement production in 2012?

- A.
24% increase

- B.
37% decrease

- C.
45% increase

- D.
60% decrease

Answer: Option C

**Explanation** :

In the given time period, there would be an increase of 18% from 2008-2009, followed by a compounded increase of 18% from 2009-2010, followed by a compounded decrease of 12% from 2010-2011. Finally, there is a compounded increase of 18%.

Hence, it is clear that effectively the production has increased. Hence, options 2 and 4 are ruled out.

Also, even if the increase is not compounded, there would have been a net increase of 18 + 18 – 12 + 18 = 42%.

Since the increase is compounded, the net effect would be more than 42%.

The exact value in 2012, if the base value is x in 2008 is x × 1.18 × 1.18 × 0.88 × 1.18 = 1.445x

This is an increase of approximately 45%.

Hence, option (c).

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**5. IIFT 2012 QA | Algebra - Logarithms**

If log 3, log (3* ^{x}* – 2) and log (3

*+ 4) are in arithmetic progression, then*

^{x}*x*is equal to

- A.
$\frac{8}{3}$

- B.
log

_{3}8 - C.
log

_{2}3 - D.
8

Answer: Option B

**Explanation** :

log(3* ^{x}* – 2) – log3 = log (3

*+ 4) – log(3*

^{x}*– 2)*

^{x}Let 3*x* = *t*

∴ log$\left\{\frac{t-2}{3}\right\}$ = log$\left\{\frac{t+4}{t-2}\right\}$

∴ $\frac{t-2}{3}$ = $\frac{t+4}{t-2}$

∴*t*^{2} + 4 – 4*t* = 3*t* + 12

∴ *t*^{2} – 7*t* – 8 = 0

∴ (3*x* – 8)(3*x* + 1) = 0

Since 3* ^{x}* cannot be negative,

∴ 3* ^{x}* = 8

∴ *x* = log_{3}8

Hence, option (b).

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**6. IIFT 2012 QA | Modern Math - Probability**

A student is required to answer 6 out of 10 questions in an examination. The questions are divided into two groups, each containing 5 questions. She is not allowed to attempt more than 4 questions from each group. The number of different ways in which the student can choose the 6 questions is

- A.
100

- B.
160

- C.
200

- D.
280

Answer: Option C

**Explanation** :

As there are 5 questions in each section we have three different cases to be considered.

**Case 1:**

3 questions from section 1 and 3 questions from section 2

Number of ways = ^{5}C_{3 }× ^{5}C_{3} = 100

**Case 2:**

4 questions from section 1 and 2 questions from section 2

Number of ways = ^{5}C_{4} × ^{5}C_{2} = 50

**Case 3:**

2 questions from section 1 and 4 questions from section 2

Number of ways = ^{5}C_{2} × ^{5}C_{4} = 50

∴ Total number of ways = 100 + 50 + 50 = 200

Hence, option (c).

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**7. IIFT 2012 QA | Modern Math - Probability**

The answer sheets of 5 engineering students can be checked by any one of 9 professors. What is the probability that all the 5 answer sheets are checked by exactly 2 professors?

- A.
$\frac{20}{2187}$

- B.
$\frac{40}{2187}$

- C.
$\frac{40}{729}$

- D.
None of the above

Answer: Option B

**Explanation** :

The paper of each student can go to any of the nine professors.

As there are 5 students, there are 9 × 9 × 9 × 9 × 9 = 9^{5} ways in which the papers can be checked by the professors.

Now, number of ways of selecting two professors = ^{9}C_{2}

The five papers can be checked by the two professors in 2^{5} ways, but this will contain two ways in which the papers are checked by just one professor.

∴ The number of ways in which 5 answer sheets are checked by exactly two professors = ^{9}C_{2} × (2^{5} – 2)

Number of ways in which each paper can be checked by a professor = 2

∴ Number of ways such that five papers can be checked by those 2 professors = ^{9}C_{2} × (2^{5} – 2)

∴ Probability = $\frac{{}^{9}{C}_{2}\times ({2}^{5}-2)}{{9}^{5}}$ = $\frac{40}{2187}$

Hence, option (b).

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**8. IIFT 2012 QA | Arithmetic - Simple & Compound Interest**

Mr. Mishra invested Rs.25,000 in two fixed deposits X and Y offering compound interest @ 6% per annum and 8% per annum respectively. If the total amount of interest accrued in two years through both fixed deposits is Rs.3518, the amount invested in Scheme X is

- A.
Rs. 12,000

- B.
Rs. 13,500

- C.
Rs. 15,000

- D.
Cannot be determined

Answer: Option C

**Explanation** :

Let the amounts invested in fixed deposits X and Y be *x* and *y* respectively.

∴ *x* + *y* = 25000 … (i)

As the interest at the end of two years is 3518,

(1.06)^{2}*x* + (1.08)^{2}*y* – 25000 = 3518

∴ 1.1236*x* + 1.1664*y* = 28518 … (ii)

Solving (i) and (ii),

*x* = 15000

Hence, option (c).

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**9. IIFT 2012 QA | Modern Math - Probability**

The probability that in a household LPG will last 60 days or more is 0.8 and that it will last at most 90 days is 0.6. The probability that the LPG will last 60 to 90 days is

- A.
0.40

- B.
0.50

- C.
0.75

- D.
None of the above

Answer: Option A

**Explanation** :

Probability that the LPG will last ≥ 60 days = 0.8

∴ Probability that the LPG will last < 60 days = 1 – 0.8 = 0.2

Probability that the LPG will last ≤ 90 days = 0.6

∴ The probability that the LPG will last ≥ 60 days and ≤ 90 days will be = (Probability that a LPG will last ≤ 90 days ) – (Probability that a LPG will last < 60 days)

= 0.6 – 0.2 = 0.4

Hence, option (a).

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**10. IIFT 2012 QA | Arithmetic - Percentage**

In 2011, Plasma – a pharmaceutical company – allocated Rs.4.5 × 10^{7} for Research and Development. In 2012, the company allocated Rs.60,000,000 for Research and Development. If each year the funds are evenly divided among 2 × 10^{2} departments, how much more will each department receive this year than it did last year?

- A.
Rs.2.0 × 10

^{5} - B.
Rs.7.5 × 10

^{5} - C.
Rs.7.5 × 10

^{4} - D.
Rs.2.5 × 10

^{7}

Answer: Option C

**Explanation** :

Funds allocated for Research & Development in 2011

= Rs. 4.5 × 10^{7}

Funds allocated for Research & Development in 2012

= Rs. 6 × 10^{7}

Difference in the funds = Rs. 1.5 × 10^{7}

= $\frac{1.5\times {10}^{7}}{2\times {10}^{2}}$ = 7.5 × 10^{4}

Hence, option (c).

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**11. IIFT 2012 QA | Geometry - Circles**

In a circular field, there is a rectangular tank of length 130 m and breadth 110 m. If the area of the land portion of the field is 20350 m2 then the radius of the field is

- A.
85 m

- B.
95 m

- C.
105 m

- D.
115 m

Answer: Option C

**Explanation** :

Area of a circular field = π*r*^{2}

Area of tank = 130 × 110 = 14300 m^{2}

∴ π*r*^{2 }– 14300 = 20350

∴ π*r*^{2 }= 34650

*r*^{2} = 1575 × 7 = 225 × 49

*r* = 15 × 7 = 105 m

Hence,** **option 3.

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**12. IIFT 2012 QA | Geometry - Circles**

A hemispherical bowl is filled with hot water to the brim. The contents of the bowl are transferred into a cylindrical vessel whose radius is 50% more than its height. If diameter of the bowl is the same as that of the vessel, the volume of the hot water in the cylindrical vessel is

- A.
60% of the cylindrical vessel

- B.
80% of the cylindrical vessel

- C.
100% of the cylindrical vessel

- D.
None of the above

Answer: Option C

**Explanation** :

Let the height of the cylindrical vessel be 2h.

∴ Radius of the cylindrical vessel is 3h.

Also the radius of the hemispherical bowl = 3h

∴ Volume of the hemispherical bowl = $\left(\frac{2}{3}\right)$ × π × (3*h*)^{3} = 18π*h*^{3}

Volume of the cylindrical vessel = π × (3*h*)^{2} × (2*h*) = 18π*h*^{3}

Hence the cylindrical vessel will be completely filled when the contents are transferred.

Hence,** **option 3.

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**13. IIFT 2012 QA | Geometry - Coordinate Geometry**

There are two buildings, one on each bank of a river, opposite to each other. From the top of one building – 60 m high, the angles of depression of the top and the foot of the other building are 30° and 60° respectively. What is the height of the other building?

- A.
30 m

- B.
18 m

- C.
40 m

- D.
20 m

Answer: Option C

**Explanation** :

Let the height of the shorter building be h.

Let the distance between the two buildings be x.

Therefore from the diagram,

tan 60 = $\frac{60}{x}$

∴ x = $\frac{60}{\sqrt{3}}$ = 20$\sqrt{3}$

∴ tan 30 = $\frac{60-h}{x}$

∴ 60 - h = 20

∴ h = 40

Hence, option (c).

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**14. IIFT 2012 QA | Arithmetic - Time, Speed & Distance**

It takes 15 seconds for a train travelling at 60 km/hour to cross entirely another train half its length and travelling in opposite direction at 48 km/hour. It also passes a bridge in 51 seconds. The length of the bridge is

- A.
550 m

- B.
450 m

- C.
500 m

- D.
600 m

Answer: Option A

**Explanation** :

Relative speed of the longer train with respect to the

shorter one = (60 + 48) × $\frac{5}{18}$ = 30 m/s

Let the length of the longer train be x m.

Now the distance travelled by the longer train to cross the shorter train = length of the longer train + length of the shorter train = x + x/2 = 3x/2

The trains cross each other with a speed of (60 + 48) km/h

(60 + 48) km/h = (60 + 48) × $\frac{5}{18}$ = 30 m/s

Therefore,

$\frac{3x}{2}$ = 30 × 15

∴ x = 300 mts

Let the length of the bridge be L.

Now total distance covered in crossing the bridge

= length of the longer train + length of the bridge

= 300 + L

∴ 300 + L = 60 × $\left(\frac{5}{18}\right)$ × 51 = 850

∴ L = 550 mts

Hence, option (a).

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**15. IIFT 2012 QA | Arithmetic - Time & Work**

12 men can complete a work in ten days. 20 women can complete the same work in twelve days. 8 men and 4 women started working and after nine days 10 more women joined them. How many days will they now take to complete the remaining work?

- A.
2 days

- B.
5 days

- C.
8 days

- D.
10 days

Answer: Option A

**Explanation** :

Let one man complete the piece of work in m days and let one woman complete the piece of work in n days.

∴ $\frac{12}{m}$ = $\frac{1}{10}$ and $\frac{20}{w}$ = $\frac{1}{12}$

∴ $\frac{1}{m}$ = $\frac{1}{120}$ and $\frac{1}{w}$ = $\frac{1}{240}$

Let n be the number of days after 9 days that the work takes to get over.

∴ 9$\left(\frac{8}{m}+\frac{4}{w}\right)$ + n$\left(\frac{8}{m}+\frac{4}{w}+\frac{10}{w}\right)$ = 1

∴ 9$\left(\frac{1}{15}+\frac{1}{60}\right)$ + n$\left(\frac{1}{15}+\frac{1}{60}+\frac{1}{24}\right)$ = 1

∴ $\frac{3}{4}$ + $\frac{15n}{120}$ = 1

∴ n = 2

Hence, option (a).

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**16. IIFT 2012 QA | Arithmetic - Time, Speed & Distance**

The Howrah-Puri express can move at 45 km/hour without its rake, and the speed is diminished by a constant that varies as the square root of the number of wagons attached. If it is known that with 9 wagons, the speed is 30 km/hour, what is the greatest number of wagons with which the train can just move?

- A.
63

- B.
64

- C.
80

- D.
81

Answer: Option C

**Explanation** :

Let the reduced speed of the train be denoted by S and the number of wagons attached to it be denoted by N.

∴ S ∝ $\sqrt{N}$

∴ $\frac{{S}_{1}}{{S}_{2}}$ = $\sqrt{\frac{{N}_{1}}{{N}_{2}}}$

The speed reduces from 45 km/hr to 30 km/hr with 9 wagons.

∴ The reduction in speed = *S*_{1} = 45 – 30 = 15 km/hr

Now, let *N*_{2} number of wagons attached when the train halts completely.

Hence, *S*_{2} = reduction in speed at this point = 45 – 0

= 45 km/hr

∴ $\frac{15}{45}$ = $\frac{\sqrt{9}}{\sqrt{{N}_{2}}}$

∴ *N _{2}* = 81 wagons

Hence, when 81 wagons are attached, the train halts completely. For the train to just move, the number of wagons attached should be 1 less than 81 i.e. 80.

Hence, option (c).

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**17. IIFT 2012 QA | Arithmetic - Average**

At a reputed Engineering College in India, total expenses of a trimester are partly fixed and partly varying linearly with the number of students. The average expense per student is Rs.400 when there are 20 students and Rs.300 when there are 40 students. When there are 80 students, what is the average expense per student?

- A.
Rs. 250

- B.
Rs. 300

- C.
Rs. 330

- D.
Rs. 350

Answer: Option A

**Explanation** :

Let the fixed expenses be k.

Let the variable expenses per student be m.

∴ $\frac{20m+k}{20}$ = 400 ...(i)

$\frac{40m+k}{40}$ = 300 ...(ii)

Solving (i) and (ii) simultaneously, we get m = 200 and k = 4000

There average cost for 80 students will be

= $\frac{80\times 200+4000}{80}$ = Rs. 250.

Hence, option (a).

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**18. IIFT 2012 QA | Arithmetic - Profit & Loss**

Rohit bought 20 soaps and 12 toothpastes. He marked-up the soaps by 15% on the cost price of each, and the toothpastes by Rs.20 on the cost price each. He sold 75% of the soaps and 8 toothpastes and made a profit of Rs.385. If the cost of a toothpaste is 60% the cost of a soap and he got no return on unsold items, what was his overall profit or loss?

- A.
Loss of Rs. 355

- B.
Loss of Rs. 210

- C.
Loss of Rs. 250

- D.
None of the above

Answer: Option A

**Explanation** :

∴ 2.25s + 160 = 385

∴ s = 100

Cost of unsold items = 5s + 4 × 0.6s = 7.4s = 740, which is a loss.

Total cost = 20s + 12 × 0.6s = 27.2s

∴ Overall loss = 740 – 385 = 355

Hence, option (a).

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**19. IIFT 2012 QA | Algebra - Simple Equations**

The value of $\sqrt{7+\sqrt{7-\sqrt{7+\sqrt{7-...\infty}}}}$ is

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option C

**Explanation** :

Let x = $\sqrt{7+\sqrt{7-\sqrt{7+\sqrt{7-...}}}}$

∴ $\sqrt{7+\sqrt{7-x}}$ = x

∴ 7 + $\sqrt{7-x}$ = x^{2}

Substituting the options, only x = 3 satisfies the equation.

Hence, option (c).

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**20. IIFT 2012 QA | Algebra - Number System**

The unit digit in the product of (8267)^{153} × (341)^{72} is

- A.
1

- B.
2

- C.
7

- D.
9

Answer: Option C

**Explanation** :

The unit’s digit of (8267)^{153} is same as unit digit of 7^{153}.

Since cyclicity of 7 is 4 and the remainder obtained when 153 is divided by 4 is 1,

∴ Unit’s digit of (8267)^{153 }= unit’s digit of 7^{153} = unit’s digit of 7^{1} = 7

Similarly,

Unit’s digit of (341)^{72} is same as unit’s digit of 1^{72} = 1

Hence the unit’s digit of the product = 7 × 1 = 7

Hence, option (c).

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**21. IIFT 2012 QA | Algebra - Number System**

Z is the product of first 31 natural numbers. If X = Z + 1, then the numbers of primes among X + 1, X + 2, ..., X + 29, X + 30 is

- A.
30

- B.
2

- C.
Cannot be determined

- D.
None of the above

Answer: Option D

**Explanation** :

Z = 31!

Z is divisible by all numbers less than 32.

X = 31! + 1

X + 1 = 31! + 2, will be divisible by 2,

X + 2 = 31! + 3 will be divisible by 3,

X + 3 = 31! + 4 will be divisible by 4 and so on.

Hence none of the numbers will be prime.

Hence, option (d).

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**22. IIFT 2012 QA | Arithmetic - Mixture, Alligation, Removal & Replacement**

A 10 litre cylinder contains a mixture of water and sugar, the volume of sugar being 15% of total volume. A few litres of the mixture is released and an equal amount of water is added. Then the same amount of the mixture as before is released and replaced with water for a second time. As a result, the sugar content becomes 10% of total volume. What is the approximate quantity of mixture released each time?

- A.
1 litres

- B.
1.2 litres

- C.
1.5 litres

- D.
2 litres

Answer: Option D

**Explanation** :

The amount of sugar in the cylinder = 1.5 litres

Now, 1.5$\left\{1-{\left(\frac{a}{10}\right)}^{2}\right\}$ is the amount of sugar

left after a litres has been replaced twice.

Now after replacement, sugar left is 10% of the total solution i.e.10% of 10 litres = 1 litre

∴ 1 = 1.5$\left\{1-{\left(\frac{a}{10}\right)}^{2}\right\}$

∴* *0.66 = $\left\{1-{\left(\frac{a}{10}\right)}^{2}\right\}$

$\sqrt{0.66}$ ≈ 0.8

∴ 0.8 = 1 - $\left(\frac{a}{10}\right)$

∴ $\left(\frac{a}{10}\right)$ = 0.2

∴ a = 2

Hence, option (d).

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**23. IIFT 2012 QA | Geometry - Circles**

Eight points lie on the circumference of a circle. The difference between the number of triangles and the number of quadrilaterals that can be formed by connecting these points is

- A.
7

- B.
14

- C.
32

- D.
84

Answer: Option B

**Explanation** :

The number of triangles formed using the 8 points

= ^{8}C_{3} = 56

The number of quadrilaterals formed using the 8 points = ^{8}C_{4 }= 70

∴ The difference = 14

Hence, option (b).

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**24. IIFT 2012 QA | Geometry - Triangles**

The perimeter of a right-angled triangle measures 234 m and the hypotenuse measures 97 m. Then the other two sides of the triangle are measured as

- A.
100 m and 37 m

- B.
72 m and 65 m

- C.
80 m and 57 m

- D.
None of the above

Answer: Option B

**Explanation** :

All the three options given satisfy the perimeter criteria. The hypotenuse is the greatest side.

∴ Option 1 is eliminated.

Now, 97^{2} = 9409

**Option 2:**

97^{2} – 72^{2} = 4225 = 65^{2}

**Option 3:**

97^{2} – 80^{2} = 3009 ≠ 57^{2}

Hence, option (b).

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**25. IIFT 2012 QA | Arithmetic - Ratio, Proportion & Variation**

A sum of Rs.1400 is divided amongst A, B, C and D such that A's share : B's share = B's share : C's share = C's share : D's share = $\frac{3}{4}$

How much is C’s share?

- A.
Rs.72

- B.
Rs.288

- C.
Rs.216

- D.
Rs.384

Answer: Option D

**Explanation** :

A : B = 3 : 4, B : C = 3 : 4, C : D = 3 : 4

∴ A : D = ${\left(\frac{3}{4}\right)}^{3}$ = $\frac{27}{64}$

Let D = 64

Then, C = $\frac{3}{4}$ × 64 = 48

B = $\frac{3}{4}$ × 48 = 36

A = $\frac{3}{4}$ × 36 = 27

∴ A : B : C : D = 27 : 36 : 48 : 64

∴ C's share = $\left\{\frac{48}{27+36+48+64}\right\}$ × 1400 = Rs. 384

Hence, option (d).

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